# Normal force changes

Tokspor
Hi guys, I am confused about how the normal force exerted on an object changes depending on the situation.

Let's say an object weighs 10 N at rest. The normal force here is 10 N as well since that is by how much the object is pushing down on the surface.

When someone tries to pull it upward with a 6 N force, it "relieves" some of the normal force. Since the object is now only pushing down on the surface with 4 N, the normal force is 4 N.

So the proper set-up for an equation here is F(normal) = F(gravity) - F(upward)

But in an elevator, let's say the elevator is accelerating upward at 2 m/s^2. If a person were inside the elevator, why doesn't this "relieve" some of their normal force as well? Instead, I have been told that the proper equation for this situation is F(upward) = F(normal) - F(gravity). Rearranging that equation, F(normal) = F(upward) + F(gravity). So now the normal force is actually much greater as a result of an upward force.

Why is this so?

cragar
action- reaction , if the elevator is accelerating upward this adds to the force , like you would feel heavier going up. And in relativity there is no difference between being accelerated by a rocket or being in a gravitational field .

Homework Helper
Welcome to PF!

Hi Tokspor! Welcome to PF! (try using the X2 tag just above the Reply box )
… So the proper set-up for an equation here is F(normal) = F(gravity) - F(upward)

But in an elevator, let's say the elevator is accelerating upward at 2 m/s^2. If a person were inside the elevator, why doesn't this "relieve" some of their normal force as well? Instead, I have been told that the proper equation for this situation is F(upward) = F(normal) - F(gravity). Rearranging that equation, F(normal) = F(upward) + F(gravity). So now the normal force is actually much greater as a result of an upward force.

Why is this so?

Because you're using "F(upward)" to mean two different things

in the first case, it's a separate (third) applied force, but in the second case it's the total of the (two) forces. Let's apply good ol' Newton's second law …

in the first case, a = 0, and so all the forces must add to 0 …

F(normal) + F(gravity) + F(applied) = 0,

ie F(normal) = mg - F(applied),​

('cos gravity is downward )

but in the second case, a = 2,

F(normal) + F(gravity) = 2,

ie F(normal) = mg + 2,​

… see? no applied force!! Mentor
So the proper set-up for an equation here is F(normal) = F(gravity) - F(upward)

But in an elevator, let's say the elevator is accelerating upward at 2 m/s^2. If a person were inside the elevator, why doesn't this "relieve" some of their normal force as well? Instead, I have been told that the proper equation for this situation is F(upward) = F(normal) - F(gravity). Rearranging that equation, F(normal) = F(upward) + F(gravity). So now the normal force is actually much greater as a result of an upward force.
The proper set-up in both cases is to use Newton's 2nd law:

ΣF = ma

F(normal) - F(gravity) = ma

When the acceleration is zero, the normal force equals your weight. If the elevator is accelerating upward (thus a > 0), the normal force is greater than your weight--you feel heavier. (The normal force must not only support your weight but accelerate you.)

Edit: While I was goofing off, tiny-tim beat me to it! sganesh88
But in an elevator, let's say the elevator is accelerating upward at 2 m/s^2. If a person were inside the elevator, why doesn't this "relieve" some of their normal force as well?

This is because in this elevator scenario, the normal force is the source of this "upward force" too. If suppose a huge guy infront holds you by your neck and lifts you, the "upward force" is provided by *his* hands. In that case, the normal force gets some help and is relieved as you hoped. (Also hope he releases you soon.)